Problem 16 - Applied Calculus

Question

A firm manufactures a commodity at two different factories. The total cost of manufacturing depends on the quantities, \(q_{1}\) and \(q_{2}\), supplied by each factory, and is expressed by the joint cost function, $$C=f\left(q_{1}, q_{2}\right)=2 q_{1}^{2}+q_{1} q_{2}+q_{2}^{2}+500$$ The company's objective is to produce 200 units, while minimizing production costs. How many units should be supplied by each factory?

Step-by-Step Solution

Answer

Factory 1 should supply 50 units, and Factory 2 should supply 150 units.

Step 1: Define the Problem

We are presented with a cost function \( C = f(q_1, q_2) = 2q_1^2 + q_1q_2 + q_2^2 + 500 \) and need to find the values of \( q_1 \) and \( q_2 \) such that the total production is 200 units, i.e., \( q_1 + q_2 = 200 \), while minimizing the cost \( C \).

Step 2: Set Up Lagrange Multipliers

To solve the optimization problem, we use Lagrange multipliers. Define the Lagrange function: \( \mathcal{L}(q_1, q_2, \lambda) = 2q_1^2 + q_1q_2 + q_2^2 + 500 + \lambda(200 - q_1 - q_2) \), where \( \lambda \) is the Lagrange multiplier for the constraint \( q_1 + q_2 = 200 \).

Step 3: Compute Partial Derivatives

Find the partial derivatives of \( \mathcal{L} \):1. \( \frac{\partial \mathcal{L}}{\partial q_1} = 4q_1 + q_2 - \lambda \)2. \( \frac{\partial \mathcal{L}}{\partial q_2} = q_1 + 2q_2 - \lambda \)3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = 200 - q_1 - q_2 \).

Step 4: Solve the System of Equations

Set each partial derivative equal to zero:1. \( 4q_1 + q_2 = \lambda \)2. \( q_1 + 2q_2 = \lambda \)3. \( q_1 + q_2 = 200 \)Solve these equations simultaneously to find \( q_1 \) and \( q_2 \).

Step 5: Calculate \( \lambda \) and Solve for \( q_1 \)

Find \( \lambda \) by equating the first two equations: \( 4q_1 + q_2 = q_1 + 2q_2 \), giving \( 3q_1 = q_2 \). Substitute \( q_2 = 3q_1 \) into the constraint: \( q_1 + 3q_1 = 200 \), so \( 4q_1 = 200 \) resulting in \( q_1 = 50 \).

Step 6: Solve for \( q_2 \)

Substitute \( q_1 = 50 \) into \( q_1 + q_2 = 200 \) to find \( q_2 = 150 \).

Step 7: Verify Solution

Verify that \( q_1 = 50 \) and \( q_2 = 150 \) satisfy both the original constraint \( q_1 + q_2 = 200 \) and minimize the cost function \( C \).

Key Concepts

Lagrange multipliers

When faced with an optimization problem with constraints, Lagrange multipliers become a valuable tool. Basically, they help find the local maxima and minima of a function subject to equality constraints. Here's how they work.

Imagine you want to minimize or maximize a function, like our cost function \( C = f(q_1, q_2) \). However, you're required to meet a certain condition, such as having the total production be 200 units. Lagrange multipliers assist by adding this constraint into the function as a new term, often involving a new variable called \( \lambda \), the Lagrange multiplier.

In our problem, the Lagrangian, denoted as \( \mathcal{L} \), includes the cost function plus \( \lambda \) times the constraint \( (200 - q_1 - q_2) \). This establishes a single equation, combining both the goal and constraint. To find optimal values, you solve the system of equations formed by the partial derivatives of \( \mathcal{L} \).

Joint cost function

The concept of a joint cost function is pivotal in understanding how different factors contribute to a total cost. In the context of our exercise, the joint cost function \( C = f(q_1, q_2) = 2q_1^2 + q_1q_2 + q_2^2 + 500 \) represents how costs are affected by the quantities produced at each of the company's two factories.

Key aspects to consider here include:

• The quadratic terms \( 2q_1^2 \) and \( q_2^2 \) suggest increasing costs at an increasing rate as production \( q_1 \) or \( q_2 \) increases. This comes from the non-linear nature of these terms.
• The term \( q_1q_2 \) implies an interaction between the two production levels, where changes in one production level impact the overall cost.
• The constant term \( 500 \) represents fixed baseline costs that do not change with production level.

Joint cost functions like this play a crucial role in business strategy, helping to balance costs across facilities and optimize production efficiently.

Partial derivatives

Partial derivatives are essential in calculus for understanding how a function changes as one variable is altered, while others are held constant. In optimization problems, they play a crucial role by providing the rates of change needed to find minima or maxima.

For the joint cost function \( C = f(q_1, q_2) \), we consider its partial derivatives to explore its behavior with respect to \( q_1 \) and \( q_2 \). Here's why it's important:

• The partial derivative \( \frac{\partial \mathcal{L}}{\partial q_1} = 4q_1 + q_2 - \lambda \) highlights how the cost changes as \( q_1 \) changes, given \( q_2 \) stays fixed.
• Similarly, \( \frac{\partial \mathcal{L}}{\partial q_2} = q_1 + 2q_2 - \lambda \) shows the sensitivity of cost with respect to changes in \( q_2 \).
• Lastly, \( \frac{\partial \mathcal{L}}{\partial \lambda} \) ensures the constraint is met by being zero when \( q_1 + q_2 = 200 \).

This way, partial derivatives guide decision-making, helping solve for optimal values by ensuring that the slopes, represented by these derivatives, balance out to zero.